https://file.aiquickdraw.com/mathbot/file/1728873648747jlc3ijhv.jpg

In the diagram, AFGB is a straight line, BCEF is a parallelogram and triangle EFG is an isosceles triangle. Find
(i) angle GFE
(ii) angle FEA

Let's solve the given problem step by step based on the information in the diagram:

### Given:
1. \( AFGB \) is a straight line.
2. \( BCEF \) is a parallelogram.
3. Triangle \( EFG \) is an isosceles triangle with \( EF = EG \).
4. \( \angle EAF = 40^\circ \) and \( \angle CEF = 65^\circ \).

#### (i) To find \( \angle GFE \):

- Since \( EFG \) is an isosceles triangle with \( EF = EG \), the base angles \( \angle GFE \) and \( \angle FEG \) are equal.
- The sum of the angles in a triangle is \( 180^\circ \), so:
  \[
  \angle GFE + \angle FEG + \angle EFG = 180^\circ
  \]
- Given that \( \angle EFG = 65^\circ \), we can set \( \angle GFE = \angle FEG = x \).
  \[
  x + x + 65^\circ = 180^\circ
  \]
  \[
  2x = 115^\circ
  \]
  \[
  x = 57.5^\circ
  \]
Therefore, \( \angle GFE = 57.5^\circ \).

#### (ii) To find \( \angle FEA \):

- Since \( AFGB \) is a straight line, the angles along this line add up to \( 180^\circ \). Therefore:
  \[
  \angle FEA + \angle EAF = 180^\circ
  \]
  \[
  \angle FEA + 40^\circ = 180^\circ
  \]
  \[
  \angle FEA = 140^\circ
  \]

### Answers:
1. \( \angle GFE = 57.5^\circ \)
2. \( \angle FEA = 140^\circ \)

Would you like more details on this, or do you have any questions?

Here are 5 related questions to consider:
1. What is the measure of \( \angle FEG \) in triangle \( EFG \)?
2. How can the properties of a parallelogram be used to find other angles in the diagram?
3. What is the measure of \( \angle BCF \) in the parallelogram \( BCEF \)?
4. How can the exterior angle theorem help solve problems in triangle geometry?
5. What properties of isosceles triangles are essential in solving angle problems?

**Tip:** Remember that the sum of angles in a triangle is always \( 180^\circ \), and this fundamental rule can help solve various geometric problems.

In the diagram, AFGB is a straight line, BCEF is a parallelogram and triangle EFG is an isosceles triangle. Find (i) angle GFE (ii) angle FEA

In this problem, we solve for angle GFE and angle FEA using geometric principles, including the isosceles triangle theorem and the properties of parallelograms. Learn how to apply the sum of angles in triangles and straight lines to find the unknown angles. Suitable for Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation